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3 years ago

8x+15 =2 (4x) im on a test please help

Mathematics

2 answers:

11Alexandr11 [23.1K]3 years ago

7 0

Answer:

Impossible

Step-by-step explanation:

the answer is 8x so you have 15 on limbo

DerKrebs [107]3 years ago

3 0

lol

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Answer:

700 or 0.76

Step-by-step explanation:

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3 years ago

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Darya [45]

Answer:

1/12 my guy

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find dy/dx by implicit differentiation. y cos x = 5x2 + 3y2

lbvjy [14]

Step 1:
Start by putting $\frac{d}{dx}$ in front of each term

$\frac{d}{dx}[y cos x]= \frac{d}{dx}[5x^2]+ \frac{d}{dx}[ 3y^2]$
-----------------------------------------------------------------------------------------------------------------
Step 2:

Deal with the terms in 'x' and the constant terms
$\frac{d}{dx}[ycosx]= 10x+ \frac{d}{dx} [3y^2]$
----------------------------------------------------------------------------------------------------------------
Step 3:

Use the chain rule for the terms in 'y'
$\frac{d}{dx}[ycosx]=10x+6y \frac{dy}{dx}$
--------------------------------------------------------------------------------------------------------------
Step 4:

Use the product rule on the term in 'x' and 'y'
$(y) \frac{d}{dx} cos x+(cos x) \frac{d}{dx}y =10x+6y \frac{dy}{dx}
$
$y(-siny)+(cosx) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
--------------------------------------------------------------------------------------------------------------

Step 5:

Rearrange to make $\frac{dy}{dx}$ the subject
$-y sin(y)+cos(x) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
$cos(x) \frac{dy}{dx}-6y \frac{dy}{dx}=10x+y sin(y)$
$[cos(x) - 6y] \frac{dy}{dx}=10x + y sin(y)$
$\frac{dy}{dx}= \frac{10x+ysin(y)}{cos(x)-6y}$ ⇒ Final Answer

5 0

4 years ago

Can someone explain this??

alexandr402 [8]

Answer:

Step-by-step explanation:

$a+(n-1)*d=t_{n}\\\\a+(70-1)*(-43)=t_{70}\\\\a+69*(-43)=500\\\\a-2967=500\\\\a=500+2967\\\\a=3467$

4 0

3 years ago

A. The solution is (−2,−3)

Cerrena [4.2K]

9514 1404 393

Answer:

D. (-3, -2)

Step-by-step explanation:

The equations have different coefficients for x and y, so will have one solution. The solutions offered are easily tested in either equation.

Using (x, y) = (-2, -3):

x = y -1 ⇒ -2 = -3 -1 . . . . False

Using (x, y) = (-3, -2):

x = y -1 ⇒ -3 = -2 -1 . . . .True

2x = 3y ⇒ 2(-3) = 3(-2) . . . . True

The solution is (-3, -2).

If you'd like to solve the set of equations, substitution for x works nicely.

2(y -1) = 3y

2y -2 = 3y . . eliminate parentheses

-2 = y . . . . . . subtract 2y

x = -2 -1 = -3

The solution is (x, y) = (-3, -2).

5 0

3 years ago